Helium is pumped into a spherical balloon at a rate of 3 cubic feet per second. How fast is the radius increasing after 3 minutes?
I know . but i don't know how to find the radius of the balloon at 3 minutes.
Note: Your hotlinked image from UGA is generating certificate errors. You might want to replace the image with the math tools provided here: $\displaystyle V\, =\, \frac{4}{3}\pi r^3$
You are given that $\displaystyle \frac{dV}{dt}\, =\, 3$. You know the "volume" formula.
Convert the "three minutes" into seconds, and find the volume after that number of seconds.
Plug this volume into the "volume" formula, and solve for the radius at that time.
Differentiate the "volume" formula with respect to time. Plug in the known values for $\displaystyle \frac{dV}{dt}$ and $\displaystyle r$, and solve for $\displaystyle \frac{dr}{dt}$.
1. The volume of a sphere is calculated by:
$\displaystyle V=\dfrac43 \pi r^3~\implies~r=\sqrt[3]{\dfrac{3V}{4\pi}}$
2. The volume after 3 min = 180 s is 540 cft.
3. Differentiate the equation for r:
$\displaystyle \dfrac{dr}{dt} = \dfrac{d\left( \sqrt[3]{\dfrac{3V}{4\pi}} \right)}{dV} \cdot \dfrac{dV}{dt}$
4. Plug in V = 540 and $\displaystyle \dfrac{dV}{dt} = 3$ to calculate the rate of change of the radius.
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